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Chapter 6 1. a. Consider an aqueous reaction of equal volumes of 2.0 M HCl and 2.0 M NAOH, with a final volume of 100.0 mL and a density of 1.03 g/ml. What will the change in heat (qrxn) be for this reaction given a positive temperature change of 2.5 degrees Celsius? Assume a value of 4.18 J g' C'for Csoln. m=dXV = l.039 x - ל to LoomL=l03 a %3D 1036(4.18) (て.sYc = lo74.355 (4.18)(て5)C - l076、355 -1,076 KJ b. Also calculate AHrxn for the acid.

Chapter 6 1. a. Consider an aqueous reaction of equal volumes of 2.0 M HCl and 2.0 M NAOH, with a final volume of 100.0 mL and a density of 1.03 g/ml. What will the change in heat (qrxn) be for this reaction given a positive temperature change of 2.5 degrees Celsius? Assume a value of 4.18 J g' C'for Csoln. m=dXV = l.039 x - ל to LoomL=l03 a %3D 1036(4.18) (て.sYc = lo74.355 (4.18)(て5)C - l076、355 -1,076 KJ b. Also calculate AHrxn for the acid.

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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I already figured out the first part of the problem, I just need help with b).

Chapter 6 1. a. Consider an aqueous reaction of equal volumes of 2.0 M HCI and 2.0 M NaOH, with a final volume of 100.0 mL and a density of 1.03 g/ml. What will the change in heat (qrxn) be for this reaction given a positive temperature change of 2.5 degrees Celsius? Assume a value of 4.18 J g' C-lfor Csoln. m=dxV = .039 l.039 x loomL=1o mL 1030(4.18) (て,5)C= l07し、355 -1076.355 -l,076KJ 7 b. Also calculate AHrxn for the acid.
Transcribed Image Text:Chapter 6 1. a. Consider an aqueous reaction of equal volumes of 2.0 M HCI and 2.0 M NaOH, with a final volume of 100.0 mL and a density of 1.03 g/ml. What will the change in heat (qrxn) be for this reaction given a positive temperature change of 2.5 degrees Celsius? Assume a value of 4.18 J g' C-lfor Csoln. m=dxV = .039 l.039 x loomL=1o mL 1030(4.18) (て,5)C= l07し、355 -1076.355 -l,076KJ 7 b. Also calculate AHrxn for the acid.
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